[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 497

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Can I have the root?
And numbers?
No root, but I like the idea. Like a cubic equation or something?
In short, it looks like we need to make a proper symmetric f-from these three variables.
P.S. Just got it. There is no root operation of any degree.
And digits... Well, if you need 5a, you could do, say, a+a+a+a+a.
Oh, it's complicated, Andrei. Where's the s?
P.S. The problem is for 8th-11th grades. No need for integrals.
No root, but I like the idea. Like a cubic equation or something?
In short, it looks like we need to make a proper symmetric f-from these three variables.
P.S. Just got it. There is no root operation of any degree.
And digits... Well, if you need 5a, you could do, say, a+a+a+a+a.
No, you have to divide it by three.
In short, good equations (x2 is the desired one)
In short, good equations (x2 -- the sought after one)
I'll add:
(a-b)*(a-c) + (b-a)*(b-c) + (c-a)*(c-b) = (x1-x2)^2 = x1^2 - 2*x1*x2 + x2^2
In short, good equations (x2 is the desired one)
something is wrong. The conditions don't mention x1 or x2.
I.e. only numbers a, b, c and arithmetic operations.
It should be:
f(a,b,c) = c
For example:
a - b + c = c
a : b * c = c
Something like this. The difficulty is that you don't know which of these 3 numbers are "the same" and which are "different", i.e. the arithmetic expression must be universal.
You can't do it without the square root, can you?
I'll add:
(a-b)*(a-c) + (b-a)*(b-c) + (c-a)*(c-b) = (x1-x2)^2 = x1^2 - 2*x1*x2 + x2^2
somehow: